This question is related to my previous question. The answers to that question inspired a new question, namely
For a complex manifold $M$, why is $H_{DR}^p(M,\mathbb{C})\cong H_{DR}^p(M,\mathbb{R})\otimes_\mathbb{R}\mathbb{C}$
Where $$H_{DR}^p(M,\mathbb{C})=\frac{\ker d:A^p(M,\mathbb{C})\to A^{p+1}(M,\mathbb{C})}{dA^{p-1}(M,\mathbb{C})}$$ And $$A^p(M,\mathbb{C})=\Gamma(\wedge^p(T^*M\otimes_\mathbb{R} \mathbb{C}_M))$$ and $\mathbb{C}_M$ is the trivial bundle over $M$ with fibre $\mathbb{C}$. And $H_{DR}^p(M,\mathbb{R})$ is the usual de Rham cohomology of the underlying real manifold, with spaces of forms $A^p(M)$. The answers to my previous question suggest that we can use the universal coefficient theorem here, but I don't see how. We have two complexes:
$$\to A^{p-1}(M)\to A^p(M)\to A^{p+1}(M)\to$$ and $$\to A^{p-1}(M,\mathbb{C})\to A^p(M,\mathbb{C})\to A^{p+1}(M,\mathbb{C})\to$$
But here it is not clear that $A_p(M)\otimes \mathbb{C}\cong A_p(M,\mathbb{C})$, since this is the statement that $$\Gamma(\wedge^p T^*M)\otimes\mathbb{C}\cong \Gamma(\wedge^p T^*M\otimes \mathbb{C})$$ which is not obvious to me. And if this is true, its a second matter of confirming that the $d$ on our complex-valued complex corresponds to $d\otimes\mathbb{1}$ under this isomorphism, although that is probably true.
In conclusion I do not see how to apply the universal coefficient theorem here. Any help on how to approach this would be much appreciated